中四Math .....一問

a)y = x^2 + bx + c

-12 = 0^2 + b ( 0 ) + c [ Sub ( 0, - 12 )]

c = -12

So,

y = x^2 + bx - 12

0 = ( - 4 )^2 + b ( - 4 ) - 12 [ Sub ( -4 , 0 )]

4b = 4

b = 1

b) y = x^2 + x - 12

0 = x^2 + x - 12

( x + 4 )( x - 3 ) = 0

x = -4 or 3

So P ( 3 , 0 ).